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\title{\heiti\zihao{2} 习题7.3}
\author{中书君}
\date{\songti 2021年1月13日}

\begin{document}
\maketitle
\section{利用定积分求极限 :}
\subsection{$\lim _{n \rightarrow \infty} \frac{1}{n^{4}}\left(1+2^{3}+\cdots+n^{3}\right)$}
\textbf{解}\quad
$$\lim _{n \rightarrow \infty} \frac{1}{n^{4}}\left(1+2^{3}+\cdots+n^{3}\right)=\lim _{n \rightarrow \infty} \sum_{i=1}^{n}\left(\frac{i}{n}\right)^{3} \cdot \frac{1}{n}=\int_{0}^{1} x^{3} \mathrm{~d} x=\frac{1}{4}$$


\subsection{$\lim _{n \rightarrow \infty} n\left[\frac{1}{(n+1)^{2}}+\frac{1}{(n+2)^{2}}+\cdots+\frac{1}{(n+n)^{2}}\right]$}
\textbf{解}\quad
$$
\begin{aligned}
    \lim _{n \rightarrow \infty} n\left[\frac{1}{(n+1)^{2}}+\cdots+\frac{1}{(n+n)^{2}}\right]&=\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{i=1}^{n} \frac{n^{2}}{(n+i)^{2}}\\&=\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{i=1}^{n} \frac{1}{\left(1+\frac{i}{n}\right)^{2}}\\&=\int_{0}^{1} \frac{1}{(1+x)^{2}} \mathrm{~d} x\\&=\frac{1}{2}
\end{aligned}
$$

\subsection{$\lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{\sin \frac{k}{n} \pi}{n+\frac{k}{n}}$}
\textbf{解}\quad
$$
\begin{array}{l}
\sum_{k=1}^{n} \frac{\sin \frac{k}{n} \pi}{n+1} \leqslant \sum_{k=1}^{n} \frac{\sin \frac{k}{n} \pi}{n+\frac{k}{n}} \leqslant \sum_{k=1}^{n} \frac{\sin \frac{k}{n} \pi}{n} \\
\lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{\sin \frac{k}{n} \pi}{n}=\int_{0}^{1} \sin \pi x \mathrm{~d} x \\
\lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{\sin \frac{k}{n} \pi}{n+1}=\lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{\sin \frac{k}{n} \pi}{n} \frac{n}{n+1}=\int_{0}^{1} \sin \pi x \mathrm{~d} x
\end{array}
$$

从而由夹逼定理,
$$
\lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{\sin \frac{k}{n} \pi}{n+\frac{k}{n}} =\int_{0}^{1} \sin \pi x \mathrm{~d} x = \frac{2}{\pi}
$$


\subsection{$\lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(1+\frac{k}{n}\right) \sin \frac{k \pi}{n^{2}}$}
\textbf{解}$1^{\circ}$\quad
对于一切 $k<n, n>3$,
$$
0 \leqslant \frac{k \pi}{n^{2}}-\sin \frac{k \pi}{n^{2}} \leqslant \tan \frac{k \pi}{n^{2}}-\sin \frac{k \pi}{n^{2}} \leqslant \tan \frac{k \pi}{n^{2}}\left(1-\cos \frac{k \pi}{n^{2}}\right) \leqslant \frac{2 k \pi}{n^{2}}\left(1-\cos \frac{\pi}{n}\right)
$$

所以
$$
0 \leqslant \sum_{k=1}^{n-1}\left(1+\frac{k}{n}\right)\left(\frac{k \pi}{n^{2}}-\sin \frac{k \pi}{n^{2}}\right) \leqslant \sum_{k=1}^{n-1}\left(1+\frac{k}{n}\right) \frac{2 k \pi}{n^{2}}\left(1-\cos \frac{\pi}{n}\right) \leqslant 2 \pi\left(1-\cos \frac{\pi}{n}\right) \rightarrow 0
$$

则有
$$
\lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(1+\frac{k}{n}\right) \sin \frac{k \pi}{n^{2}}=\lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(1+\frac{k}{n}\right) \frac{k \pi}{n^{2}}=\int_{0}^{1}(1+x) \pi x \mathrm{~d} x=\frac{5 \pi}{6}
$$

\textbf{解}$2^{\circ}$\quad
首先把 $\sin \frac{k \pi}{n^{2}}$ 泰勒展开成 $\frac{k \pi}{n^{2}}+O\left(\frac{k^{3} \pi^{3}}{n^{6}}\right)$
然后代入:
$$
\begin{aligned}
    \lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(1+\frac{k}{n}\right) \sin \frac{k \pi}{n^{2}}&=\lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(1+\frac{k}{n}\right)\left(\frac{k \pi}{n^{2}}+O\left(\frac{k^{3} \pi^{3}}{n^{6}}\right)\right)
\end{aligned}
$$
又因为
$$
\lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(1+\frac{k}{n}\right) O\left(\frac{k^{3} \pi^{3}}{n^{6}}\right)<\lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(1+\frac{n}{n}\right) O\left(\frac{n^{3} \pi^{3}}{n^{6}}\right)=0
$$
所以
$$
\lim_{n \rightarrow \infty} \sum_{k=1}^{n}\left(1+\frac{k}{n}\right)\left(\frac{k \pi}{n^{2}}+O\left(\frac{k^{3} \pi^{3}}{n^{6}}\right)\right)=\lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(1+\frac{k}{n}\right)\left(\frac{k \pi}{n^{2}}\right)
$$
则有
$$
\lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(1+\frac{k}{n}\right) \sin \frac{k \pi}{n^{2}}=\lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(1+\frac{k}{n}\right) \frac{k \pi}{n^{2}}=\int_{0}^{1}(1+x) \pi x \mathrm{~d} x=\frac{5 \pi}{6}
$$

\section{求下列极限:}
\subsection{$\lim _{x \rightarrow 0} \frac{1}{x} \int_{0}^{x} \cos t^{2} \mathrm{~d} t$}
\textbf{解}\quad
$$\frac{\mathrm{~d}}{\mathrm{~d} x} \int_{0}^{x} \cos ^{2} t \mathrm{~d} t=\cos ^{2} x$$

由洛必达法则:
$$\lim _{x \rightarrow 0} \frac{\int_{0}^{x} \cos ^{2} t \mathrm{~d} t}{x}=\frac{\cos ^{2} 0}{1}=1$$

\subsection{$\lim _{x \rightarrow \infty} \frac{\left(\int_{0}^{x} \mathrm{e}^{t^{2}} \mathrm{~d} t\right)^{2}}{\int_{0}^{x} \mathrm{e}^{2 t^{2}} \mathrm{~d} t}$}
\textbf{解}\quad
$$
\lim _{x \rightarrow \infty} \frac{\left(\int_{0}^{x} \mathrm{e}^{t^{2}} \mathrm{~d} t\right)^{2}}{\int_{0}^{x} \mathrm{e}^{2 t^{2}} \mathrm{~d} t}=\lim _{x \rightarrow \infty} \frac{2 \mathrm{e}^{x^{2}} \int_{0}^{x} \mathrm{e}^{t^{2}} \mathrm{~d} t}{\mathrm{e}^{2 x^{2}}}=\lim _{x \rightarrow \infty} \frac{2 \mathrm{e}^{x^{2}}}{2 x \mathrm{e}^{x^{2}}}=0 .
$$

\section{计算下列定积分:}
\subsection{$\int_{1}^{4} \frac{1}{x(1+\sqrt{x})} \mathrm{d} x$}
\textbf{解}\quad
令 $u=\sqrt{x}$,则$$ \int_{1}^{4} \frac{1}{x(1+\sqrt{x})} \mathrm{d} x=\int_{1}^{2} \frac{2 u}{u^{2}(1+u)} \mathrm{d} u=2 \int_{1}^{2}\left(\frac{1}{u}-\frac{1}{1+u}\right) \mathrm{d} u=2 \ln \frac{4}{3}$$


\subsection{$\int_{1 / 4}^{1 / 2} \frac{\arcsin \sqrt{x}}{\sqrt{x(1-x)}} \mathrm{d} x$}
\textbf{解}\quad
令$u = \sqrt{x}$,$v=\arcsin u$,则原式:$$\int_{\frac{1}{4}}^{\frac{1}{2}} \frac{\arcsin \sqrt{x}}{\sqrt{x(1-x)}} \mathrm{~d} x=\int_{\frac{1}{2}}^{\sqrt{\frac{1}{2}}} \frac{2 \arcsin u}{\sqrt{1-u^{2}}} \mathrm{~d} u=2 \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} v \mathrm{~d} v=\left.v^{2}\right|_{\frac{\pi}{6}} ^{\frac{\pi}{4}}=\frac{5\pi^{2}}{144}$$


\subsection{$\int_{0}^{1} \frac{1}{\left(x^{2}-x+1\right)^{\frac{3}{2}}} \mathrm{~d} x$}
\textbf{解}\quad
令$u = x-\frac{1}{2},u=\frac{\sqrt{3}}{2} \tan v$,则原式:$$\int_{0}^{1} \frac{1}{\left(x^{2}-x+1\right)^{\frac{3}{2}}} \mathrm{~d} x=\int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{8}{\left(4 u^{2}+3\right)^{\frac{3}{2}}} \mathrm{~d} u=\frac{4}{3} \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \cos v \mathrm{~d} v=\frac{4}{3}$$


\subsection{$\int_{0}^{\pi / 2} \frac{\cos x}{\sin x+\cos x} \mathrm{~d} x$}
\textbf{解}\quad
令$u = \frac{\pi}{2}-x$则原式:$$\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\sin x+\cos x} \mathrm{~d} x =-\int_{\frac{\pi}{2}}^{0} \frac{\cos \left(\frac{\pi}{2}-t\right)}{\sin \left(\frac{\pi}{2}-t\right)+\cos \left(\frac{\pi}{2}-t\right)} \mathrm{~d} t =\int_{0}^{\frac{\pi}{2}} \frac{\sin t}{\cos t+\sin t} \mathrm{~d} t$$

又因为:$$\int_{0}^{\frac{\pi}{2}} \frac{\sin t+\cos t}{\cos t+\sin t} \mathrm{~d} t=\frac{\pi}{2}$$

从而原式$=\frac{\pi}{4}$


\subsection{$\int_{-1}^{1} \frac{x \mathrm{~d} x}{\sqrt{5-4 x}}$}
\textbf{解}\quad
令$u = \sqrt{5-4x}$
$$\int_{-1}^{1} \frac{x}{\sqrt{5-4 x}} \mathrm{~d} x=\frac{1}{8} \int_{3}^{1} u^{2}-5 \mathrm{~d} u=\left.\frac{1}{8}\left(\frac{1}{3} u^{3}-5 u\right)\right|_{3} ^{1}=\frac{1}{6}$$


\subsection{$\int_{0}^{a} x^{2} \sqrt{a^{2}-x^{2}} \mathrm{~d} x$}
\textbf{解}\quad
令$x = a\sin u$,则原式:
$$\int_{0}^{a} x^{2} \sqrt{a^{2}-x^{2}} \mathrm{~d} x=a^{4} \int_{0}^{\frac{\pi}{2}} \sin ^{2} u \cos ^{2} u \mathrm{~d} u=a^{4} \int_{0}^{\frac{\pi}{2}} \frac{1-\cos 4u}{8} \mathrm{~d} u=\frac{\pi}{16} a^{4}$$


\subsection{$\int_{0}^{\ln 2} \sqrt{\mathrm{e}^{x}-1} \mathrm{~d} x$}
\textbf{解}\quad
令$u = e^{x},v = \sqrt{u-1}$,则原式:
$$\int_{0}^{\ln 2} \sqrt{\mathrm{e}^{x}-1} \mathrm{~d} x=\int_{1}^{2} \frac{\sqrt{u-1}}{u} \mathrm{~d} u=2 \int_{0}^{1} \frac{v^{2}}{v^{2}+1} \mathrm{~d} v=2-\frac{\pi}{2}$$


\subsection{$\int_{-2}^{-1} \frac{1}{x \sqrt{x^{2}-1}} \mathrm{~d} x$}
\textbf{解}\quad
令$u = \sqrt{x^{2} - 1}$,则原式:
$$\int_{-2}^{-1} \frac{1}{x \sqrt{x^{2}-1}} \mathrm{~d} x =\int_{\sqrt{3}}^{0} \frac{1}{u^{2}+1} \mathrm{~d} u=-\frac{\pi}{3}$$


\subsection{$\int_{0}^{\pi} \frac{x \sin x}{1+\cos ^{2} x} \mathrm{~d} x$}
\textbf{解}\quad
由$$\int_{0}^{\pi} x f(\sin x) \mathrm{d} x=\frac{\pi}{2} \int_{0}^{\pi} f(\sin x) \mathrm{d} x$$

从而
$$\begin{aligned} \int_{0}^{\pi} \frac{x \sin x}{1+\cos ^{2} x} \mathrm{~d} x &=\frac{\pi}{2} \int_{0}^{\pi} \frac{\sin x}{1+\cos ^{2} x} \mathrm{~d} x=-\frac{\pi}{2} \int_{0}^{\pi} \frac{1}{1+\cos ^{2} x} \mathrm{~d}(\cos x) \\ &=-\left.\frac{\pi}{2}[\arctan (\cos x)]\right|_{0} ^{\pi}=\frac{\pi^{2}}{4} \end{aligned}$$


\subsection{$\int_{0}^{2} \frac{1+x^{2}}{1+x^{4}} \mathrm{~d} x$}
\textbf{解}\quad
$$\begin{aligned}\int_{0}^{2} \frac{1+x^{2}}{1+x^{4}} \mathrm{~d} x&=\lim _{a \rightarrow 0} \int_{a}^{2} \frac{1+x^{2}}{1+x^{4}} \mathrm{~d} x=\left.\lim _{a \rightarrow 0} \frac{1}{\sqrt{2}} \arctan \frac{x^{2}-1}{\sqrt{2} x}\right|_{a} ^{2} \\&=\lim _{a \rightarrow 0} \frac{1}{\sqrt{2}} \arctan \frac{x^{2}-1}{\sqrt{2} x}\\&=\frac{1}{\sqrt{2}} \arctan \frac{3}{2 \sqrt{2}}+\frac{\pi}{2 \sqrt{2}}\end{aligned}$$

\section{计算定积分:}
\subsection{$\int_{0}^{2} \max \left\{x^{2}, x\right\} \mathrm{d} x$}
\textbf{解}\quad
$$
\int_{0}^{2} \max \left\{x^{2}, x\right\} \mathrm{d} x=\int_{0}^{1} x \mathrm{~d} x+\int_{1}^{2} x^{2} \mathrm{~d} x=\frac{17}{6}
$$


\subsection{$\int_{-1}^{1} \frac{2 x^{2}+x}{1+\sqrt{1-x^{2}}} \mathrm{~d} x$}
\textbf{解}\quad
$$
\begin{aligned}
\int_{-1}^{1} \frac{2 x^{2}+x}{1+\sqrt{1-x^{2}}} \mathrm{~d} x &=\int_{-1}^{1} \frac{2 x^{2}}{1+\sqrt{1-x^{2}}} \mathrm{~d} x=4 \int_{0}^{1} \frac{x^{2}}{1+\sqrt{1-x^{2}}} \mathrm{~d} x \\
&=4 \int_{0}^{\frac{\pi}{2}}(1-\cos t) \mathrm{d}(\sin t)=4-\pi
\end{aligned}
$$


\subsection{$\int_{0}^{\pi} \sqrt{\sin ^{3} x-\sin ^{5} x} \mathrm{~d} x$}
\textbf{解}\quad
$$
\begin{aligned}
\int_{0}^{\pi} \sqrt{\sin ^{3} x-\sin ^{5} x} \mathrm{~d} x &=\int_{0}^{\pi} \sqrt{\sin ^{3} x\left(1-\sin ^{2} x\right)}=\int_{0}^{\pi} \sqrt{\sin ^{3} x} \cdot \cos x \mathrm{~d} x \\
&=\int_{0}^{\pi} \sqrt{\sin ^{3} x} \mathrm{~d}(\sin x)=\int_{0}^{1} \sqrt{t^{3} } \mathrm{~d}t = \frac{2}{5}
\end{aligned}
$$


\subsection{$\int_{1 / \mathrm{e}}^{e}|\ln x| \mathrm{d} x$}
\textbf{解}\quad
$$
\int_{\frac{1}{e}}^{e}|\ln x| \mathrm{~d} x=-\int_{\frac{1}{e}}^{1} \ln x \mathrm{~d} x+\int_{1}^{e} \ln x \mathrm{~d} x=-\int_{-1}^{0} t e^{t} \mathrm{~d} t+\int_{0}^{1} t e^{t} \mathrm{~d} t=2-\frac{e}{2}
$$


\subsection{$\int_{0}^{2}(2 x+1) \sqrt{2 x-x^{2}} \mathrm{~d} x$}
\textbf{解}\quad
$$
\begin{aligned}
\int_{0}^{2}(2 x+1) \sqrt{2 x-x^{2}} \mathrm{~d} x&=-\int_{0}^{2} \sqrt{2 x-x^{2}} d\left(2 x-x^{2}\right)+3 \int_{0}^{2} \sqrt{2 x-x^{2}} \mathrm{~d} x=0+3 \int_{0}^{2} \sqrt{2 x-x^{2}} \mathrm{~d} x\\&=
3 \int_{0}^{2} \sqrt{1-(x-1)^{2}} \mathrm{~d} x=3 \int_{-1}^{1} \sqrt{1-t^{2}} \mathrm{~d} t= 6 \int_{0}^{1} \sqrt{1-t^{2}} \mathrm{~d} t\\&
=6 \int_{0}^{\frac{\pi}{2}} \cos ^{2} u \mathrm{~d} u=-\frac{3 \pi}{2}
\end{aligned}
$$


\subsection{$\int_{0}^{1} x(\arctan x)^{2} \mathrm{~d} x$}
\textbf{解}\quad
$$
\begin{aligned}
\int_{0}^{1} x(\arctan x)^{2} \mathrm{~d} x &=\left.\frac{1}{2} x^{2}(\arctan x)^{2}\right|_{0} ^{1}-\int_{0}^{1} \frac{x^{2} \arctan x}{1+x^{2}} \mathrm{~d} x \\
&=\frac{\pi^{2}}{32}-\int_{0}^{1}\left(\arctan x-\frac{\arctan x}{1+x^{2}}\right) \mathrm{d} x \\
&=\frac{\pi^{2}}{32}-\left.x \arctan x\right|_{0} ^{1}-\int_{0}^{1} \frac{x}{1+x^{2}} \mathrm{~d} x+\left.\frac{1}{2}(\arctan x)^{2}\right|_{0} ^{1} \\
&=\frac{\pi^{2}}{16}-\frac{\pi}{4}-\frac{1}{2} \ln 2
\end{aligned}
$$


\subsection{$\int_{0}^{1} \frac{\ln (1+x)}{1+x^{2}} \mathrm{~d} x$}
\textbf{解}\quad
做变换 $x=\tan t, \mathrm{~d} x=\sec ^{2} t \mathrm{~d} t$
$$
\int_{0}^{1} \frac{\ln (1+x)}{1+x^{2}} \mathrm{~d} x=\int_{0}^{\frac{\pi}{4}} \ln (1+\tan t) \mathrm{d} t
$$

由
$$
\ln \left[1+\tan \left(\frac{\pi}{4}-t\right)\right]=\ln \left(1+\frac{1-\tan t}{1+\tan t}\right)=\ln \frac{2}{1+\tan t}=\ln 2-\ln (1+\tan t)
$$

知 $\ln (1+\tan t)-\frac{1}{2} \ln 2$ 是在区间 $\left[0, \frac{\pi}{4}\right]$ 上关于区间中点的奇函数,即
$$
\int_{0}^{\frac{\pi}{4}}\left[\ln (1+\tan t)-\frac{1}{2} \ln 2\right] \mathrm{d} t=0
$$

所以
$$
\int_{0}^{\frac{\pi}{4}} \ln (1+\tan t) \mathrm{d} t=\int_{0}^{\frac{\pi}{4}} \frac{1}{2} \ln 2 \mathrm{~d} t=\frac{\pi}{8} \ln 2
$$


\subsection{$\int_{0}^{\ln 2} x \mathrm{e}^{-x} \mathrm{~d} x$}
\textbf{解}\quad
令$e^x=u$,则原式:
$$
\quad \int_{0}^{\ln 2} x \mathrm{e}^{-x} \mathrm{~d} x=\int_{1}^{2} \ln t \cdot \frac{1}{t} \mathrm{~d} t=\frac{1}{2} \int_{0}^{1} \mathrm{~d}\left(\ln ^{2} t\right)=(\ln 2)^{2}
$$


\subsection{$\int_{0}^{1} x^{2} \sqrt{1-x^{2}} \mathrm{~d} x$}
\textbf{解}\quad
由3.6知,其为$\frac{\pi}{16}$


\subsection{10) $\int_{0}^{2} \frac{x \sqrt{4-x^{2}}}{\sqrt{4+x^{2}}} \mathrm{~d} x$}
\textbf{解}\quad
$$
\begin{aligned} \int_{0}^{2} \frac{x \sqrt{4-x^{2}}}{\sqrt{4+x^{2}}} \mathrm{~d} x &=\frac{1}{2} \int_{0}^{2} \frac{\sqrt{4-x^{2}}}{\sqrt{4+x^{2}}} \mathrm{~d} x^{2}=\frac{1}{2} \int_{0}^{4} \frac{\sqrt{4-u}}{\sqrt{4+u}} \mathrm{~d} u \\ &=\frac{1}{2} \int_{0}^{4} \frac{4-u}{\sqrt{16-u^{2}}} \mathrm{~d} u \\ &=2 \int_{0}^{4} \frac{1}{\sqrt{16-u^{2}}} \mathrm{~d} u-\frac{1}{2} \int_{0}^{4} \frac{u}{\sqrt{16-u^{2}}} \mathrm{~d} u \\ &=\left.\left(2\arcsin \frac{u}{4}+\frac{1}{2} \sqrt{16-u^{2}}\right)\right|_{0} ^{4}=\pi-2 \end{aligned}
$$

\section{函数 $f(x)$ 在 $[0,+\infty)$ 上连续, 且 $\lim _{x \rightarrow+\infty} f(x)=a .$ \\证明 $\lim _{x \rightarrow+\infty} \frac{1}{x} \int_{0}^{x} f(t) \mathrm{d} t=a$.}
\textbf{证}\quad
$$
\lim _{x \rightarrow+\infty} \frac{\int_{0}^{x} f(t) \mathrm{~d} t}{x}=\frac{\lim_{x \rightarrow+\infty} f(x)}{1}=a
$$

\section{设函数 $f(x)$ 在[0,1]上连续,且 $f(x)>0,$ 试将下面极限表示为积分形式$$\lim _{n \rightarrow \infty} \sqrt[n]{f\left(\frac{1}{n}\right) f\left(\frac{2}{n}\right) \cdots f\left(\frac{n-1}{n}\right) f\left(\frac{n}{n}\right)}$$}
\textbf{解}\quad
$$
\mathrm{LHS}=\lim _{n \rightarrow+\infty} e^{\sum_{i=1}^{n} f\left(\frac{i}{n}\right) \cdot \frac{1}{n}}=e^{\int_{0}^{1} f(x) \mathrm{~d} x}
$$

\section{设 $f(x)$ 在 $[a, b]$ 上连续,且单调增加,证明:$$\int_{a}^{b} x f(x) \mathrm{d} x \geqslant \frac{a+b}{2} \int_{a}^{b} f(x) \mathrm{d} x$$}
\textbf{证}\quad
设函数
$$
F(x)=\int_{a}^{x} t f(t) \mathrm{~d} t-\frac{a+x}{2} \int_{a}^{x} f(t) \mathrm{~d} t, x \in[a, b]
$$

则原题目等价于证明 $F(b) \geq 0=F(a) .$ 如果函数 $F(x)$ 单调递增,则结论自然成立,下面证明函
数 $F(x)$ 单调增加.
$$
\begin{aligned}
F^{\prime}(x) &=x f(x)-\frac{1}{2} \int_{a}^{x} f(t) \mathrm{~d} t-\frac{a+x}{2} f(x) \\
&=\frac{x-a}{2} f(x)-\frac{1}{2} \int_{a}^{x} f(t) \mathrm{~d} t=\frac{1}{2} \int_{a}^{x}[f(x) \mathrm{~d} t-f(t) \mathrm{~d} t] \\
&=\frac{1}{2} \int_{a}^{x}[f(x)-f(t)] \mathrm{~d} t
\end{aligned}
$$

因为 $x \geq t,$ 函数 $f(x)$ 单调递增,所以 $f(x) \geq f(t),$ 由积分的保号性可知
$$
F^{\prime}(x) \geq 0
$$

因此
$$
F(b) \geq F(a)=0
$$

\section{若函数 $f(x)$ 连续可导,且 $f(0)=0, f(1)=1 .$ 求证$$\int_{0}^{1}\left|f(x)-f^{\prime}(x)\right| \mathrm{d} x \geqslant \frac{1}{\mathrm{e}}$$}
\textbf{证}\quad
令 $F(x)=\mathrm{e}^{-x} f(x),$ 则 $F(0)=0, F(1)=\frac{1}{\mathrm{e}}$

$$\frac{1}{\mathrm{e}}=F(\mathrm{e})-F(1)=\int_{0}^{1} F^{\prime}(x) \mathrm{d} x=\int_{0}^{1} \mathrm{e}^{-x}\left[-f(x)+f^{\prime}(x)\right] \mathrm{d} x \leqslant \int_{0}^{1}\left|f(x)-f^{\prime}(x)\right| \mathrm{d} x$$

\section{证明 $: \int_{0}^{\frac{\pi}{2}}(\sin \theta-\cos \theta) \ln (\sin \theta+\cos \theta) \mathrm{d} \theta=0$}
\textbf{证}\quad
$$
\sin \theta+\cos \theta=\sqrt{2} \sin \left(\theta+\frac{\pi}{4}\right)=\sqrt{2} \cos \left(\frac{\pi}{4}-\theta\right)=\sqrt{2} \cos \left(\theta-\frac{\pi}{4}\right)\\
\sin \theta-\cos \theta=\sqrt{2} \sin \left(\theta-\frac{\pi}{4}\right)
$$

可见,$(\sin \theta - \cos  \theta)$是奇函数,$(\sin \theta + \cos  \theta)$是偶函数.

从而原式被积式为关于$(\frac{\pi}{4},0)$中心对称的奇函数,从而原式积分为$0$

\section{设 $f(x)$ 在 $[a, b]$ 上连续,且满足条件 :$$\int_{a}^{b} x^{k} f(x) \mathrm{d} x=0,(k=0,1, \cdots, n)$$证明函数 $f(x)$ 在 $(a, b)$ 内至少有 $n+1$ 个不同的零点.}
\textbf{证}$1^{\circ}$\quad
用数学归纳法,对于 $k=0,$ 由 $\int_{a}^{b} f(x) \mathrm{d} x=0$ 和 $f(x)$ 在 $[a, b]$ 上连续知 $, f(x)$ 在
$[a, b]$ 上变号或恒为零,因此结论显然成立. 设对于 $k=n-1$ 结论成立,令
$$
F(x)=\int_{a}^{x} f(t) \mathrm{d} t
$$

显然 $F(a)=F(b)=0,$ 且有
$$
0=\int_{a}^{b} x^{k} f(x) \mathrm{d} x=\int_{a}^{b} x^{k} \mathrm{~d} F(x)=\left.x^{k} F(x)\right|_{a} ^{-b}-\int_{a}^{b} k x^{k-1} F(x) \mathrm{d} x=-\int_{a}^{b} k x^{k-1} F(x) \mathrm{d} x
$$

由 $\int_{a}^{b} x^{k} f(x) \mathrm{d} x=0 \Rightarrow \int_{a}^{b} x^{k-1} F(x) \mathrm{d} x=0$ 对 $k=1,2,3, \cdots, n,$ 都成立.
对 $F(x)$ 使用归纳法,知其在 $(a, b)$ 内至少有 $n$ 个不同的零点,因此在 $[a, b]$ 上有 $n+2$ 不 同零点,由罗尔中值定理知 $f(x)$ 在 $(a, b)$ 内至少有 $n+1$ 个不同的零点.

\textbf{证}$2^{\circ}$\quad
用反证法:

设有一个 $f$ 满足所有题设条件,但其零点个数不超过 $n .$ 这时 $f$ 在 $[a, b]$ 内的任何闭子区间上不会恒等于 $0,$ 因此从条件 $\int_{a}^{b} f(x) \mathrm{d} x=0$ 可见 $f$ 在区间 $[a, b]$ 上一定变号. 利用 $f$ 的所有零点或其中的一部分, 可以作出区间 $[a, b]$ 的一个分划
$$
P=\left\{x_{0}, x_{1}, \cdots, x_{k}\right\}
$$

其中 $x_{0}=a, x_{k}=b,$ 中间的 $k-1$ 个分点都是 $f$ 的零点, $f$ 在每个子区间上不变号.
由反证法前提可见 $k-1 \leqslant n$.
对于分划 $P$ 可以构造辅助多项式
$$
g(x)=\left(x-x_{1}\right)^{\delta_{1}}\left(x-x_{2}\right)^{\delta_{2}} \cdots\left(x-x_{k-1}\right)^{\delta_{k-1}}
$$

其中$\delta_{i}$满足若在$x_{i}$点左右函数变号则为$1$,否则为$0$

$g$ 关于分划 $P$ 具有与 $f$ 相同的性质: 即在每个子区间上不变号, 而在相邻的子区间上符号的变化一致. 这样就知道 $f \cdot g$ 在整个区间 $[a, b]$ 上不变号, 因此
$$
\int_{a}^{b} f(x) g(x) \mathrm{d} x>0(<0)
$$

另一方面, 由于 $g$ 是次数不超过 $n$ 的多项式, 从题设条件可见上述积分应当等于 $0,$ 从而引出矛盾.

\section{补充说明:}
对
$$
\int\frac{1+x^2}{1+x^4}dx=\int \frac{1+\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}} d x=\int \frac{1}{\left(x-\frac{1}{x}\right)+2} d\left(x-\frac{1}{x}\right)=\frac{1}{\sqrt{2}} \arctan \frac{x-\frac{1}{x}}{\sqrt{2}}+C
$$

$\mathrm{RHS}$并不是原函数,其在$x=0$处不连续,从而不可导,从而不能说$F^{\prime}(0)=f(0)$.

解决方法:

补充定义,使其成为一个连续函数.但$\mathrm{RHS}$对于$0$来说是一个跳跃间断点.需要非常规的补充定义手法.
$$
F(x)=\left\{\begin{array}{ll}
\frac{1}{\sqrt{2}} \arctan \frac{x-\frac{1}{x}}{\sqrt{2}}+\frac{\pi}{2} \cdot \frac{1}{\sqrt{2}} &, x>0 \\
0 & ,  x=0 \\
\frac{1}{\sqrt{2}} \arctan \frac{x-\frac{1}{x}}{\sqrt{2}}-\frac{\pi}{2} \cdot \frac{1}{\sqrt{2}} &, x<0
\end{array}\right.
$$
从而使得其为一个连续函数.所以此时的$F$是$f$的一个原函数.

但当我们算定积分的时候,其可视为一个瑕积分,将间断点直接舍弃即可.


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